Solve -2x^2=20x+50

$$-2 { x }^{ 2 } =20x+50$$

Solve for x

$$x=-5$$

Steps Using Factoring By Grouping

Subtract 20x from both sides.

$$-2x^{2}-20x=50$$

Subtract 50 from both sides.

$$-2x^{2}-20x-50=0$$

Divide both sides by 2.

$$-x^{2}-10x-25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-25. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-\left(-25\right)=25$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.

$$-1,-25 -5,-5$$

Calculate the sum for each pair.

$$-1-25=-26 -5-5=-10$$

The solution is the pair that gives sum -10.

$$a=-5 b=-5$$

Rewrite -x^{2}-10x-25 as \left(-x^{2}-5x\right)+\left(-5x-25\right).

$$\left(-x^{2}-5x\right)+\left(-5x-25\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(-x-5\right)+5\left(-x-5\right)$$

Factor out common term -x-5 by using distributive property.

$$\left(-x-5\right)\left(x+5\right)$$

To find equation solutions, solve -x-5=0 and x+5=0.

$$x=-5 x=-5$$

Steps Using the Quadratic Formula

Subtract 20x from both sides.

$$-2x^{2}-20x=50$$

Subtract 50 from both sides.

$$-2x^{2}-20x-50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -20 for b, and -50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-2\right)\left(-50\right)}}{2\left(-2\right)}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-2\right)\left(-50\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-20\right)±\sqrt{400+8\left(-50\right)}}{2\left(-2\right)}$$

Multiply 8 times -50.

$$x=\frac{-\left(-20\right)±\sqrt{400-400}}{2\left(-2\right)}$$

Add 400 to -400.

$$x=\frac{-\left(-20\right)±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

$$x=-\frac{-20}{2\left(-2\right)}$$

The opposite of -20 is 20.

$$x=\frac{20}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{20}{-4}$$

Divide 20 by -4.

$$x=-5$$

Steps for Completing the Square

Subtract 20x from both sides.

$$-2x^{2}-20x=50$$

Divide both sides by -2.

$$\frac{-2x^{2}-20x}{-2}=\frac{50}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{20}{-2}\right)x=\frac{50}{-2}$$

Divide -20 by -2.

$$x^{2}+10x=\frac{50}{-2}$$

Divide 50 by -2.

$$x^{2}+10x=-25$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-25+5^{2}$$

Square 5.

$$x^{2}+10x+25=-25+25$$

Add -25 to 25.

$$x^{2}+10x+25=0$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+5=0 x+5=0$$

Subtract 5 from both sides of the equation.

$$x=-5 x=-5$$

The equation is now solved. Solutions are the same.

$$x=-5$$